Question

2^2х + 14·2^х + 1 – 29 = 0

Answer 1

2^(2x)+28·2^(x)-29=0
2^(x)=y>0
y²+28y-29=0
y1=1 или у2=-29<0 (не уд ОДЗ)
2^(x)=1⇒x=0

Answer 2

To solve the equation 2^2x + 14 * 2^x + 1 - 29 = 0, you can use a substitution to make it easier to solve. Let's make a substitution to simplify the equation.

Let y = 2^x.

Now, the equation becomes:

y^2 + 14y + 1 - 29 = 0

This is a quadratic equation in terms of y. To solve it, we can rearrange it:

y^2 + 14y - 28 = 0

Now, we can solve this quadratic equation. You can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 14, and c = -28. Plug these values into the quadratic formula:

y = (-14 ± √(14^2 - 4 * 1 * (-28))) / (2 * 1)

y = (-14 ± √(196 + 112)) / 2

y = (-14 ± √308) / 2

Now, you have two potential values for y.

1. y₁ = (-14 + √308) / 2
2. y₂ = (-14 - √308) / 2

Now, we need to convert these values back to x using the original substitution:

1. For y₁: 2^x = (-14 + √308) / 2
2. For y₂: 2^x = (-14 - √308) / 2

To find the values of x, take the natural logarithm (ln) of both sides:

1. For y₁: x = ln((-14 + √308) / 2)
2. For y₂: x = ln((-14 - √308) / 2)

You can now calculate the approximate values of x using a calculator.